The value of $sin(2cos^(-1)(1/2))$ is what?

Question

when I plug it into the calculator (in radians), the answer is $sqrt(3)/2$ but the answer is supposed to be $1/3$ and i don't understand why

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Answer 1 · 2018-06-02T05:07:56

$sin 2 arccos(1/2) = pm sqrt{3}/2$

It doesn't matter if it's done in degrees or radians.

We'll treat the inverse cosine as multivalued. Of course a cosine of $1/2$ is one of the two tired triangles of trig.

$arccos(1/2) = pm 60^circ + 360^circ k quad$ integer $k$

Double that, $2 arccos(1/2) = pm 120^circ$

So $sin 2 arccos(1/2) = pm sqrt{3}/2$

Even when the question writers don't have to use 30/60/90 they do. But let's do

$sin 2 arccos (a/b)$

We have $sin(2a) = 2 sin a cos a$ so

$sin 2 arccos(a/b) = 2 sin arccos (a/b) cos arccos(a/b)$

$sin 2 arccos(a/b) = {2a}/b sin arccos(a/b)$

If the cosine is $a/b$ that's a right triangle with adjacent $a$ and hypotenuse $b$ , so opposite $pm sqrt{b^2-a^2}.$

$sin 2 arccos(a/b) = {2a}/b cdot (pm sqrt{b^2-a^2})/b$

$sin 2 arccos(a/b) = pm {2a}/b^2 sqrt{b^2-a^2}$

In this problem we have $a=1 and b=2$ so

$sin 2 arccos(1/2) = pm 1/2 sqrt{3} quad sqrt$

The principal value is positive.